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Home , Differential form, Divergence, Dot product, Exterior algebra, Exterior calculus, Exterior derivative

VISUALIZING EXTERIOR

GREGORY BIXLER

Abstract. , broadly, is the structure of differential forms. These are usually presented algebraically, or as computational tools to simplify Stokes’ Theorem. But geometric objects like forms should be visualized! Here, I present visualizations of forms that attempt to capture their .

Contents Introduction 1 1. Exterior 2 1.1. Vectors and dual vectors 2 1.2. The wedge of two vectors 3 1.3. Defining the 4 1.4. The wedge of two dual vectors 5 1.5. R3 has no mixed wedges 6 1.6. Interior 7 2. Integration 8 2.1. Differential forms 9 2.2. Visualizing differential forms 9 2.3. Integrating differential forms over 9 3. Differentiation 11 3.1. Exterior 11 3.2. Visualizing the 13 3.3. Closed and exact forms 14 3.4. Future : 15 Acknowledgments 16 References 16

Introduction What’s the natural way to integrate over a smooth ? Well, a k-dimensional manifold is (locally) parametrized by k coordinate functions. At each point we get k vectors, which together form an infinitesimal k-dimensional ele- ment. And how should we this element? It seems reasonable to ask for a f taking k vectors, so that • f is linear in each vector • f is zero for degenerate elements (i.e. when vectors are linearly dependent) • f is smooth These are the properties of a differential form. 1 2 GREGORY BIXLER

Introducing structure to a manifold usually follows a certain pattern: • Create it in Rn. We define the exterior algebra in 1. • Introduce it locally in coordinate neighborhoods. We devote sections 2 and 3 to this. • Understand it globally by patching together the coordinate neighborhoods. This is beyond the scope of this paper, except for a few remarks at the end of section 3.

1. Exterior algebra Given a vector space1 V , we can construct its exterior algebra2 Λ(V ), a se- quence of vector spaces Λk(V ) which represent k-dimensional volume elements.We measure these volume elements via the Λk(V )∨. This generalizes the line seen in a typical introductory class: R F · dr. Here, “F ·” is a dual vector measuring the line element dr. We face a double challenge: to see the exterior algebra of dual vectors. We’ll tackle this by first visualizing dual vectors, and then visualizing the exterior algebra of (regular) vectors and dual vectors in tandem.

1.1. Vectors and dual vectors. Visually, a vector in Rn is an arrow whose tail is at the origin. It’s less clear how to visualize dual vectors. Since there is an isomorphism V ∨ =∼ V , why not simply visualize V ∨ as V itself? The main problem with this approach is that it transforms the wrong way under a change of . For example, suppose we’re initially measuring in feet and switch to inches: if our old basis is {e1, . . . , en}, then our new basis is {e1/12, . . . , en/12}. Vectors appear to get 12 times bigger. But for any T ∈ V ∨ and v ∈ V , the value of T (v) is independent of basis, so T must shrink by a factor of 12, not grow. Instead, in R1, view a dual vector a collection of “tick marks” on the axis:

In Rn, we can do the same, with a caveat. We must choose an axis with tick marks, and project the vector into this axis.

1 In this paper, all vector spaces are finite-dimensional. 2 The notation “Λ” seems to be a corruption of the symbol ∧, the exterior/wedge . Some authors write V(V ) instead. VISUALIZING EXTERIOR CALCULUS 3

This doesn’t really measure the vector: it measures the projection. We can measure vectors more directly by orthogonally extending the tick marks. This yields a of lines in R2, planes in R3, and hyperplanes in general.

Formally, we can think of this geometric representation of a dual vector T as the set of pre-images T −1(n) of integers n.

1.2. The wedge of two vectors. Allow me to start with a historical note. In R3, in addition to the operations of addition and scaling, we have the and . We compute the dot product u · v by projecting u along v and then multiplying their signed lengths. We compute the cross product u × v by first forming the parallelogram P spanned by u and v and then setting the length of u × v to be the of P , in the direction of the (oriented) to P . These operations were not always called “dot product” and “cross product”. Hermann Grassman essentially invented in his 1844 book Die lineale Ausdehnungslehre ([3]), including the exterior algebra. In the preface, he calls these the “”3 and “exterior product,” respectively. He reasons that the dot product takes its maximum value when one vector is inside the span of the other, while the cross product is only nonzero if each vector is outside the span of the other.

3 This is not the same as the interior multiplication we define later! Also, this is probably where the term “inner product” comes from. 4 GREGORY BIXLER

Anyways, back to describing multi-dimensional volume. The cross product is essentially the right idea—it’s no accident that it shows up whenever we compute a . To generalize it, we need to fix a quirk of the cross product: it describes a 2-dimensional area as a 1-dimensional vector. The fix is to describe the result as a “2-dimensional vector”!4 We implement this fix by defining a new , the wedge product (or exterior product) u ∧ v of vectors u and v. We want it to satisfy many of the same properties as the cross product:

• (u1 + u2) ∧ v = u1 ∧ v + u2 ∧ v • c(u ∧ v) = (cu) ∧ v = u ∧ (cv) • u ∧ v = −(v ∧ u), so in particular, v ∧ v = 0

1.3. Defining the exterior algebra. The wedge product looks essentially like the product, but with some extra relations (in the third bullet above). So, it should seem natural to define the space of 2-vectors, Λ2(V ), as a quotient of the tensor V ⊗ V . Before we do this, recall a little trick, which shows that the relations v ∧ v = 0 imply u ∧ v = −v ∧ u. 0 = (u + v) ∧ (u + v) = u ∧ u + u ∧ v + v ∧ u + v ∧ v = u ∧ v + v ∧ u. We now define V ⊗ V Λ2(V ) = , I where I is the subspace of V ⊗ V generated by the v ⊗ v. We can generalize this definition to higher powers: Definition 1.1. Let V be a vector space, and k a nonnegative integer. Then V ⊗k Λk(V ) = , Ik ⊗k where Ik is the subspace of V generated by tensors v1 ⊗ · · · ⊗ vk, where at least one of the vi is repeated. (Or equivalently, where the vi are linearly dependent.)

Recall that, if {e1, . . . , en} is a basis for V , then the elements

eI = ei1 ∧ · · · ∧ eik form a basis for Λk(V ). Here, I ranges over all strictly increasing k-tuples in k n {1, . . . , n}. The of Λ (V ) is the of such tuples, which is k . Finally, treating the exterior algebra as a whole makes the definition even simpler: Let T (V ) be the for V . Then Λ(V ) = T (V )/I, where I is the generated by the tensors v ⊗ v.5

4 3 Perhaps I’m being unfair to the cross product here. It really is remarkable that in R , we can describe area using vectors, and the cross product is particularly useful since it spits out an object of the same type as its inputs. 5 The ideal I is different from the subspace I from the definition of Λ2(V ), since now we also use ⊗ when generating I. VISUALIZING EXTERIOR CALCULUS 5

1.4. The wedge of two dual vectors. We’ve seen that the wedge of two vectors looks like a stretchy, oriented parallelogram. How about the wedge of two dual vectors? First, recall the usual definition of 2-forms.

Proposition 1.2. Let V be a finite-dimensional vector space. Then

Λ2(V ∨) = W,

6 where W is the space of bilinear, alternating functions V × V → R.

Let’s start simply, in R2. Here, we have a way to measure oriented parallelo- grams: the . Indeed, the function

(u, v) 7→ det u v is bilinear and alternating. The determinant in R2 is a particular 2-form; recall n ∨ ∨  ∨ that in general, the determinant detn in R is equal to e1 ∧ · · · ∧ en , where ej is the usual basis for (Rn)∨. Here’s how we can visualize the determinant: Place a dot at each lattice point in the . To approximate det(u ∧ v), draw the parallelogram u ∧ v and count how many dots are inside. Keep track of orientations with a small circle around each dot. For example, in the below figure, det(u1 ∧ u2) ≈ 3 and det(v1 ∧ v2) ≈ −4.

From now on, I’ll mostly ignore orientations. To rescale the determinant, move the dots closer together or farther apart, reversing the circles if the is negative. Due to the bilinearity of the determinant, it doesn’t matter how you choose to move the dots.

6 A function is alternating if its sign flips when swapping any two of its arguments. 6 GREGORY BIXLER

In R3, we can get a 2-form by first choosing a plane and placing dots, projecting the 2-vector onto the plane, and counting the dots in the projection. Like with dual vectors, we can un-project the dots, which gives us a collection of evenly spaced lines.

To summarize: we can geometrically represent the wedge of two dual vectors −1 −1 α ∧ β as α (Z) ∩ β (Z). In general, we represent a (pure) k-form ω1 ∧ · · · ∧ ωk as −1 −1 ω1 (Z) ∩ · · · ∩ ωk (Z). 1.5. R3 has no mixed wedges. There’s an important distinction between pure and mixed tensors. Pure tensors are those that can be written as a single term v1 ⊗ · · · ⊗ vk, whereas mixed tensors cannot. For example, while you can write

u1 ⊗ v1 + u1 ⊗ v2 + u2 ⊗ v1 + u2 ⊗ v2 = (u1 + u2) ⊗ (v1 + v2), you typically cannot find vectors a and b so that

u1 ⊗ v1 + u2 ⊗ v2 = a ⊗ b. As long as the of V and W are at least 2, we can pick linearly indepen- dent vectors u1, u2 ∈ U and v1, v2 ∈ V . Then, U ⊗ V will have mixed tensors, such k as u1 ⊗ v1 + u2 ⊗ v2. The same, however, is not always true for Λ (V ). VISUALIZING EXTERIOR CALCULUS 7

Proposition 1.3. Let V have dimension n. Then Λk(V ) has mixed wedges if and only if 1 < k < n − 1. This may explain why differential forms developed much later than classical multivariable calculus. In three or fewer dimensions, we never need to worry about mixed wedges, so we can integrate over lines and surfaces using regular vectors. Even in higher dimensions, we can integrate over curves and hypersurfaces with im- punity. We only needed forms when we wanted to be able to integrate 2-dimensional surfaces in R4, for example. Proof sketch of (⇐). Consider the function7 J : V k → Λk(V ) given by

J(v1, . . . , vk) = v1 ∧ · · · ∧ vk. Then Λk(V ) has mixed tensors if and only if J is not surjective. The function J is not linear, but it is smooth. (Λk(V ) is a smooth manifold n since it is a finite-dimensional real vector space.) The derivative of J is a k × nk matrix8, which has the form DJ = A | −A | · · · | (−1)k+1A n because the wedge product is alternating. Since A is a k × n , (DJ) = rank(A) ≤ n. n k When 1 < k < n − 1, we have rank(DJ) ≤ n < k , so every point of V is a critical point. Hence, by Sard’s theorem, J cannot be surjective.  Proof sketch of (⇒). The space Λn(V ) has dimension 1, so all wedges are parallel to some pure wedge, and are hence pure. When k = 1, the space Λk(V ) is just V itself and J is the identity map. Let P be the set of pure (n−1)-wedges. We show that P is closed under addition. There is an isomorphism ⊥: P → V , given by taking the normal vector, scaled by volume. (In R3, ⊥ maps u∧v to u×v.) Though many can have the same normal vector, they’re all equivalent when considered as , so ⊥ is −1 injective. Add pure wedges by adding their normal vectors and applying ⊥ .  1.6. Interior multiplication. Instead of thinking of a k-form as a Λk(V ) → R, we may think of it as an alternating k-linear function V k → R. (In other words, Λk(V ∨) =∼ Λk(V )∨.) Hence, we can define an operation ι: V × Λk(V ∨) → Λk−1(V ∨) by plugging a specified vector v into the first argument of a form: ιvω = ω(v, −). Here, I’m omitting the free variables: the above definition means

(ιv(ω))(v1, . . . , vk−1) = ω(v, v1, . . . , vk−1). We call ι interior multiplication.9

7 I chose J for join, since J combines vectors. The more obvious choice, W for wedge, could be confused for another vector space. 8 Here, I’m picking a basis for V and using the corresponding basis for Λk(V ). 9 The operation ι has many names: interior product, interior derivative, interior multiplication, contraction, insertion, and probably several more. It is “interior” because it generalizes the inner product: a 1-form ω can be represented via ω = a· for some vector a, so

ιvω = ω(v) = a · v. 8 GREGORY BIXLER

We visualize interior multiplication as follows: From a k-form ω, a density of (n − k)-dimensional spaces, we obtain a (k − 1) form ιxω of (n − k + 1)-dimensional spaces by extending the spaces along direction x and rescaling properly.

In the above illustration, “Π” is the that we’re evaluating ω on. Here’s the reasoning behind this visualization. Consider a pure k-form ω in Rn and a vector v. (We restrict to pure forms since they are easier to visualize.) There are two cases:

• Suppose ιvω is not identically zero. Then we can find ω2, . . . , ωk so that

ω = ω1 ∧ · · · ∧ ωk,

where ω is the dual vector to v (here, I mean ω (w) = 1 v · w). Now, 1 1 kvk2 we have

ιvω = ω2 ∧ · · · ∧ ωk. Then ω is represented visually by

−1 −1 −1  ω1 (Z) ∩ ω2 (Z) ∩ · · · ∩ ωk (Z) , | {z } visual representation of ιv ω

−1 so we visually get from ω to ιvω by “un-intersecting” with ω1 (Z). In other words, take each subspace making up the density of ω and span it with v. • If ιvω is identically zero, we cheat a little: v is already “in” ω, so ιvω looks the same as ω. Now, ιvω is being considered as a (k − 1)-form, but has (n − k)-planes instead of (n − k + 1)-planes, so is degenerate.

2. Integration We now understand multivectors and exterior forms in , so we can let them vary and do calculus to them. In this section, we describe how to put exterior algebra on manifolds, and how to integrate over manifolds.

The reason ι is sometimes called a “derivative” is that it is an anti-derivation; in other words, if ω is a k-form, then k ιv(ω ∧ η) = ιv(ω) ∧ η + (−1) ω ∧ (ιvη). VISUALIZING EXTERIOR CALCULUS 9

2.1. Differential forms. A differential k-form ω associates to each point p in a k smooth manifold M an exterior k-form ωp ∈ Λ (TpM), so that ω “varies smoothly”. What does it mean to vary smoothly? n In open submanifolds of R , we can write ωp uniquely as X ωp = aI (p) dxI , I inc

Then, “varying smoothly” means the functions aI (p) are smooth. The key here n is that open submanifolds of R have a global coordinate chart, so the dxI make sense everywhere. To make sense of varying smoothly on other smooth manifolds, we need the notion of a , specifically the k-th exterior bundle. For details about this construction, see [4, p. 359]. The space of all differential k-forms on a manifold is often denoted Ωk(M). One mnemonic is that the shape “Ω” is a smoothed-out version of the shape “Λ”.

2.2. Visualizing differential forms. We visualize differential forms like how we visualize vector fields. To visualize vector fields, we pick several points in M. At each point p, draw the vector corresponding to p, using p as the origin. To visualize differential forms, just replace “draw the vector” with “draw the exterior form”. For example, below is the differential form ω = dx + xy dy.

2.3. Integrating differential forms over manifolds. We rigorously define the integral of a differential form as follows: (1) If ω is a differential n-form in Rn, we know ω = a(x) det for some smooth function a : Rn → R. We then define Z Z ω = a, K K where K ⊂ Rn is compact, and the RHS is the usual . (2) If ω is a differential k-form in a n-dimensional manifold M, and K is a com- pact set contained inside a coordinate neighborhood g(U) with parametriza- tion g : U → M, then we pull back ω to a differential form g∗ω over U and define Z Z ω = g∗ω, K g−1(K) where the RHS integral was defined in the first step. 10 GREGORY BIXLER

(3) Integrating over a noncompact set A is fairly technical, and involves par- titions of unity. Essentially, this involves breaking A into many compact sets Ki, possibly with overlap, and then integrating over each Ki while accounting for the overlap.

Pullback looks like projecting the linear spaces in ω into the to M, and then sending it backwards to U via g.

While is crucial to defining these integrals, we can ignore it when vi- sualizing them. Visually, this is how we integrate a differential n-form ω over a n-dimensional manifold M:

(1) Break M into tiny n-dimensional pieces Ki (2) Approximate each Ki with a tangent n-vector vi with proper (3) In each Ki, approximate ω with an exterior n-form ωi (4) Evaluate ωi(vi), and add them all up

We can take a shortcut. Since the Ki are tiny, they are almost identical to the vi. So, we just need to give an orientation to each Ki and evaluate “ωi(Ki)”, in the same way we would visually evaluate ωi(vi): by counting how many pieces of ωi pass through Ki. The image below shows the process with this shortcut. VISUALIZING EXTERIOR CALCULUS 11

In some cases, we can streamline the visualization even more. For example, let 10 f : R2 → R be smooth, and consider its form ω ∈ Ω1(R2), defined as

ωp(v) = v · (gradf)(p), where gradf is the gradient of f. In this case, we can (imprecisely) describe ω as a linear approximation to the level sets f −1(Z).

Now, to integrate ω over a curve c, we can simply see how many times c passes R through the level sets of f. In the below figure, c ω ≈ (−3) + 5 = 2.

3. Differentiation In single-variable calculus, there is essentially one derivative you can define— you can interpret the derivative of a function in many ways, but each yields the same object. However, in exterior calculus, some of those interpretations diverge, giving different mathematical objects. In this section, we describe the main one: the exterior derivative.

3.1. Exterior Derivative. We can define the derivative of a function f : R → R via the fundamental theorem of calculus Z b f(b) − f(a) = f 0(x) dx . a Pick some x ∈ R. When the [a, b] is small, we can approximate Z b f 0(x) dx ≈ (b − a)f 0(x), a

10 The form ω is actually the exterior derivative of f, which we’ll define later. This explains why you need a Riemannian metric to define a gradient vector field: the gradient is essentially a 1-form, and you need an inner product for the isomorphism V =∼ V ∨. 12 GREGORY BIXLER so f(b) − f(a) f 0(x) ≈ . b − a This is essentially the same as the difference quotient. However, the similarity ends when we step up in dimension. Now, we can think of flux and . Let F be a smooth vector field in R3, and B be a closed unit ball centered at p with boundary S. The states Z Z flux of F over S = divF. S B When the ball B is small, we can approximate 1 Z divF ≈ flux of F over S volB S and, in fact, could define divergence as 1 Z divF = lim (flux of F over S). →0 vol(B) S We can interpret the fundamental theorem of calculus as a “divergence theorem” by thinking of f(b) as the amount of f leaving [a, b] at b, and −f(a) as the amount of f leaving [a, b] at a. The negative sign indicates that at a, f leaves to the left. Using this flux-and-divergence interpretation of the derivative for inspiration, we can define the exterior derivative. (This discussion and definition was adapted from [1, p. 188].)

Definition 3.1. Let ω be a k-form, and let v1, . . . , vk+1 be vectors. For all  > 0, let Π be the (oriented) parallelepiped spanned by the vectors v1, . . . , vk+1. The exterior derivative of ω, denoted dω, is defined by 1 Z dω (v1, . . . , vk+1) = lim ω. →0 vol(Π) ∂(Π) In other words, dω is defined to make the equation Z Z dω = ω Π ∂Π true for tiny parallelepipeds Π. But we use tiny parallelepipeds to approximate manifolds, so the Generalized Stokes’ Theorem shouldn’t be a surprise. Theorem 3.2 (Generalized Stokes’ Theorem). Let M be an oriented manifold with boundary ∂M, and ω be a differential k-form on ∂M. Then Z Z dω = ω. M ∂M With our definition of the exterior derivative, the proof of Stokes’ Theorem is conceptually straightforward: (1) Show it holds when M is a parallelepiped (2) By collapsing parts of the boundary of a parallelepiped, show it holds when M is a simplex, and therefore a (3) Approximate M by polyhedra. (If M has no global coordinate chart, then use a partition of unity.) Show the formula holds in the . VISUALIZING EXTERIOR CALCULUS 13

The trickiest part of proving Stokes’ Theorem is showing that our definition is well-defined (i.e. the limit exists and produces a k-form), and is equivalent to the usual definition. I won’t prove this, but here’s an example illustrating a particular case. In the figure below, we’re evaluating dωp (εe1, εe2), where ω = a dx + b dy is 2 2 a 1-form in R and {e1, e2} is the for R . Based on the picture,      ∂b ∂a dω (εe , εe ) = −εa(p + εe ) + εa(p) + εb(p + εe ) − εb(p) ≈ ε ε − ε . p 1 2 2 1 ∂x ∂y This agrees with the usual definition, since  ∂b ∂a dω = − dx ∧ dy . ∂x ∂y The same sort of pattern happens in higher dimensions: opposite faces have opposite orientations, so we’re essentially computing partial . The orientation of the parallelepiped gives us the correct signs on the partial derivatives.

3.2. Visualizing the exterior derivative. We can use the ideas of flux and divergence to visualize the exterior derivative. This is clearest with 1-forms in R2. Consider the below 1-form ω, which we’ve drawn along a small loop centered about a point p. The flux of ω about this loop is roughly 3 + 2 − 1 − 2 = 2, so ω has divergence 2 near p. To visualize this, consider trying to “stitch up” the lines of ω like in the below right. There’s no way to do this perfectly; there will always be two “loose threads”. Hence, we can visualize dω at p with two dots. 14 GREGORY BIXLER

3.3. Closed and exact forms. Recall that a form ω is exact if ω = dη for some η, and closed if dω = 0. Since d2 = 0, all exact forms are closed. The converse is only partially true: by the Poincar´eLemma, all closed forms are locally exact. That is, all points have a neighborhood on which the form is exact.11 Hence, we can interpret the exterior derivative as a local barrier to exactness. To do this, select several points, and then try to stitch up ω as before. There is a difference: previously, we selected several points around a single central point, whereas now we just scatter some points. For example, consider the form dy. Since d(dy) = 0, we expect that dy could be stitched up. Indeed, we can simply extend the horizontal lines to meet:

If we try to stitch up x dy in the same way, we end up with loose threads popping up at each lattice point of the plane. The set of lattice points is precisely our visual representation of dx ∧ dy, which is d(x dy).

One might object: why not stitch up x dy differently, like the following?

11We could continue down this path: by checking which locally exact forms are not globally exact, we can probe the of our manifold. This leads to the de Rham . VISUALIZING EXTERIOR CALCULUS 15

The problem is that in the middle, the stitching doesn’t agree with the form—it’s curved, but the form is horizontal. But wasn’t the correct stitching also curved? It looks like it was, but it wasn’t. The below image shows, from left to right, (1) what the form x dy actually looks like (2) what the correct patching looks like (3) what the incorrect patching looks like

So, a form isn’t closed if stitching it together requires loose threads. What about a form that is closed, but not exact? The typical example is the form dθ = (y dx − x dy)/(x2 + y2) defined on R2 \ 0. As you can see below, dθ has many loose threads starting from the origin, which is outside the domain of dθ. (The stitching looks unnatural because I used a small number of points. With more points, dθ should look like many spokes sticking out of the origin.)

3.4. Future work: Lie derivative. The exterior derivative generalizes flux and divergence. If we instead generalize the “difference quotient”, we end up with the , and then the Lie derivative Lvω of a k-form ω along a path given by the vector field v. Though the Lie derivative and exterior derivative are quite different ways of generalizing the derivative to exterior calculus, they are related by a simple formula, called Cartan’s magic formula, which also bring in interior multiplication.

Lv = dιv + ιvd. To prove this, one usually shows the LHS and RHS are a special type of called a derivation, which implies it need only be shown for 0-forms and 1-forms. This proof gives an interesting view of the of these operators. 16 GREGORY BIXLER

However, it certainly isn’t visual. I spent much of the summer hunting for com- patible visualizations of interior multiplication and the Lie and exterior derivatives, hoping that all would fit in the same picture and shine a new light on this formula. Such a picture remains elusive. Maybe you can find it! Arnold hints at a visual proof with a different style of visualization in [1, p. 198]. Acknowledgments. It is a pleasure to thank my mentor, Kevin Casto, for guid- ing me through this project. I would also like to thank Aygul Galimova, Alex Manchester, and Doug Stryker for several fruitful conversations. I especially thank William Garland for helping me think through several of my visualization attempts. Finally, I thank Peter May for organizing the wonderful U Chicago REU program.

References [1] V. I. Arnold. Mathematical Methods of , 2nd edition. Springer-Verlag. 1989. [2] David Bachman. A Geometric Approach to Differential Forms. https://arxiv.org/abs/math/0306194v1 [3] Henri Grassman. Die lineale Ausdehnungslehre. [4] John M. Lee. Introduction to Smooth Manifolds, 2nd edition. Springer-Verlag. 2013. [5] Dan Piponi. On the Visualization of Differential Forms. http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.126.1099&rep=rep1&type=pdf [6] Gabriel Weinreich. Geometrical Vectors. Chicago University Press. 1998.